SOLUTION: The word problem is: Susan invests twice as much money at 7.5% as she does at 6%. If her total interest after a year is $840, how much does she have invested at each rate? Th

Question 277625: The word problem is:
Susan invests twice as much money at 7.5% as she does at 6%. If her total interest after a year is $840, how much does she have invested at each rate?
The only examples in the book and from my teacher provide a total amount invested and I'm not sure where to go without that information. I can do the part combining the system of equations without any issue but I don't see how to get the proper equations from that word problem.
I keep coming up with something like 2X+L=840 and another equation like .075X+.06L=840 but I know that can't be right.
I really appreciate whoever will be able to help me! :)
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Let her invest $x at 6%
She will invest $2x at 7.5%
the sum of individual interests = $840
0.06*x+0.075*2x=840
0.06x+0.15x=840
0.21x=840
x=840/0.21
=4000 at 6%
She invests $ 8000 at 7.5%

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